If for a non - zero x, 3x2 + 5x + 3 = 0, then the value of $${x^3} + \frac{1}{{{x^3}}}$$ is?
A. $$\frac{{10}}{{27}}$$
B. $$ - \left( {\frac{{10}}{{27}}} \right)$$
C. $$\frac{2}{3}$$
D. $$ - \left( {\frac{2}{3}} \right)$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & 3{x^2} + 5x + 3 = 0 \cr & 3{x^2} + 3 = - 5x \cr & {\text{Divide by }}3x{\text{ both sides}} \cr & x + \frac{1}{x} = \frac{{ - 5}}{3} \cr & {\text{Then, }}{x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\frac{{ - 5}}{3}} \right)^3} - 3 \times \left( {\frac{{ - 5}}{3}} \right) \cr & = \frac{{ - 125}}{{27}} + 5 \cr & = \frac{{10}}{{27}} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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